1. Day 151 – Arguments about volume of common solids

2. Introduction
We understand that objects are not always made in the geometric forms that we handle in our class rooms. For instance, most spheres that we meet in life are not perfect spheres. In cases where a cylinder is used in real world, one may find that it is a slanting sphere and not an upright sphere as most books put it. Due to this reason, we have to come up with a way to determining the volume of such shapes. This is where Cavalieri comes in. In this lesson, we are going to give an informal argument using Cavalieri's principle for the formulas for the volume of a sphere and other solid figures.

3. Vocabulary
Cylinder It is solid with a curved surface and a cross-section area that is a circle Sphere It is a solid whose shape is that of a ball. That is, all points on its surface are at equal distance from its center. Pyramid It is solid with a base and curved or triangular surfaces along its height having a common apex.

4. Vocabulary
Cone It is a pyramid whose base is a circle Prism It is a solid with a polygonal base which is also its cross section and rectangles along its height, or length.

5. Volume of solids Cavalieri‘s developed a result that is used to estimate the volume of figures that looks deformed from their standard shapes. We will use that principle to determine the volume of solids. The Cavalieri‘s principle for solids says that if two solids are enclosed between two parallel planes, and at any given point, a parallel plane is drawn on which the cross-section area of the figures on that plane are equal, then the figures have the same volume.

6. Volume of solids We will illustrate this using a cylinder
Volume of solids We will illustrate this using a cylinder. Consider cylinder A and B between parallel planes, 𝑃 1 and 𝑃 2 . ℎ 2𝑟
𝑃 2
𝑃 1

7. When we introduce another plane, 𝑃 3 parallel to both 𝑃 1 and 𝑃 2 and the cross sections if A and B on 𝑃 3 , are equal, 𝑐 1 = 𝑐 2 , then the cylinders have the same volume, 𝑉=𝜋 𝑟 2 ℎ.
𝑃 2 ℎ 2𝑟
𝑐 1
𝑐 2
𝑃 3
𝑃 1

8. Volume of a cone and other pyramids
Volume of a cone and other pyramids. The volume of the a pyramid is given by 1 3 ℎ×𝐵𝐴, where 𝐵𝐴 is the base area. Using Cavalieri‘s principle, we have the area of the following figures equal since they have the same size of the cross section at any plane parallel to the two plains at the base and top.

9. Volume of a cone and other pyramids
Volume of a cone and other pyramids. The volume of the a pyramid is given by 1 3 ℎ×𝐵𝐴, where 𝐵𝐴 is the base area. Since the base of the cone is a circle, its volume will be 1 3 ℎ×𝐵𝐴= 1 3 ℎ×𝜋 𝑟 2

10. Volume of a sphere We would like to find the volume of a sphere using the Cavalieri‘s principle. We consider a sphere inside a cylinder whose diameter is equal to its height as well as the diameter of the sphere. We would also like to have a cone dug inside it as shown then use the setup to find the volume of the sphere using the Cavalieri‘s principle.

11. If we have any plane say, 0.5𝑟 from the center, we have We determine the half chord AB using Pythagorean theorem. Its size is 𝐴𝐵= 𝑟 2 − 0.5𝑟 2 =𝑟 1−0.25 =𝑟 0.75 Thus the radius of the circle of the cross section at the sphere is 𝑟 0.75 units
0.5r r 2r A B

12. We now determine the area of the cross section in the cylinder
We now determine the area of the cross section in the cylinder. Its area is 𝜋 𝑟 2 =0.75 𝜋𝑟 2 If we consider the cross section. We know that G is 0.5𝑟 from the center, O. FG is the radius of the small hollow in the cross section. E F D G O

13. FE and FG are perpendicular to ED hence parallel to each other and triangle FGD and FED are similar. 𝐸𝐷=2𝑟,hence 𝑂𝐸=𝑟. Since 𝑂𝐺=0.5𝑟, 𝐺𝐷=0.5𝑟 By proportionality of the sides, we have 𝐺𝐷 𝐸𝐷 = 𝐹𝐺 𝐹𝐸 Upon substitution, we get 0.5𝑟 𝑟 = 𝐹𝐺 𝑟 𝐹𝐺=0.5𝑟 Thus, the radius of the smaller circle at the intersection is equal is 0.25𝑟 while that of the bigger circle is 𝑟, hence the area of the cross sectional ring is 𝜋 𝑅 2 −𝜋 𝑟 2 =𝜋( 𝑅 2 − 𝑟 2 )

14. 𝜋 𝑅 2 −𝜋 𝑟 2 =𝜋 𝑅 2 − 𝑟 2 =𝜋 𝑟 2 − 0. 5𝑟 2 =𝜋 𝑟 2 −0. 25 𝑟 2 =0
𝜋 𝑅 2 −𝜋 𝑟 2 =𝜋 𝑅 2 − 𝑟 2 =𝜋 𝑟 2 − 0.5𝑟 2 =𝜋 𝑟 2 −0.25 𝑟 2 =0.75𝜋 𝑟 2 Thus, the area of the two cross sections are all equal hence the Cavalieri‘s principle applies. Thus, the volume of the sphere is equal to the volume of the cylinder with the conical shaped hollow. Volume of sphere = Volume of the cylinder – volume of the cone =𝜋 𝑟 2 ℎ − 1 3 𝜋 𝑟 2 ℎ But the height, ℎ=2𝑟 Upon substitution, we get V=𝜋 𝑟 2 2𝑟 − 1 3 𝜋 𝑟 2 2𝑟 =2𝜋 𝑟 3 − 2 3 𝜋 𝑟 3 = 4 3 𝜋 𝑟 3

15. Example A sphere and a cylinder with a conical hollow are enclosed between two parallel planes. The diameter of the sphere is equal to that of the cylinder which is 23 𝑖𝑛. If another plane parallel to the two above cuts through their two solids, the cross section are equal. Find the volume of the sphere using Cavalieri‘s principle. Solution Since the cross sections on an intermediate plane are equal, the solids have the same volume by Cavalieri‘s principle. Radius = height, ℎ=11.5 𝑖𝑛 Volume of the cone is 1 3 𝜋 𝑟 2 ℎ= 1 3 𝜋 =1593 in 3

16. Volume of the cylinder is 𝜋 𝑟 2 ℎ=𝜋 11
Volume of the cylinder is 𝜋 𝑟 2 ℎ=𝜋 =4778 in 3 Volume of the sphere is 3185 𝑖 𝑛 3 .

17. homework
A sphere and a cylinder with a conical hollow are enclosed between two parallel planes. The radius of the sphere is equal to that of the cylinder which is 8 𝑖𝑛. If another plane parallel to the two above cuts through their two solids, the cross section are equal. Find the volume of the sphere using Cavalieri‘s principle. Write the answer to the nearest unit.

18. Answers to homework
2145 𝑖 𝑛 3

19. THE END